%!TEX program = xelatex
%!TEX TS-program = xelatex
%!TEX encoding = UTF-8 Unicode

\documentclass[12pt,t,aspectratio=169,mathserif]{beamer}
%Other possible values are: 1610, 149, 54, 43 and 32. By default, it is to 128mm by 96mm(4:3).
%run XeLaTeX to compile.

\input{wang-slides-preamble.tex}

\begin{document}

\title{高等代数二}
\subtitle{9-3-正定二次型 }
%\institute{上海立信会计金融学院}
%\author{王立庆}
\author{{\ppr LQW}}
\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
%\date{{\ppr 2023年3月9日} }

\maketitle

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{9.3.i. 作业：星期天晚上十点半之前在网络教学平台提交 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{enumerate}
\item   整理课堂笔记，补充没写完的计算或证明。
\item   习题(9.3)\#1,2,3,4,5, 抄写题目。
\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{9.3.ii. 目录 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{enumerate}

\item[9.3.1.] 正定二次型和正定矩阵的定义
\item[9.3.3.] 半正定二次型和半正定矩阵的定义
\item[9.3.4.] 定理9.3.1. 实二次型是正定的当且仅当它的秩与符号差都是$n$
\item[9.3.5.] 顺序主子式的定义
\item[9.3.6.] 定理9.3.2. 实二次型是正定的当且仅当所有顺序主子式都大于零 
\item[9.3.8.] 证明必要性
\item[9.3.10.] 证明充分性

\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{9.3.iii. 课堂讲解重点 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{enumerate}
\item  正定二次型的概念
\item  正定矩阵的概念
\item  正定二次型与正定矩阵的判别条件
\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{9.3.1. 正定二次型和正定矩阵的定义}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  {\color{red}问题：什么是正定二次型？什么是正定矩阵？}

\item  解答：
\begin{enumerate}
\item  实二次型 $q(x_1,\cdots,x_n)$ 称为是正定二次型，如果对任意不全为零的实数 $x_1,\cdots,x_n$, 这个二次型的值总是正的，也就是说，
$$\forall (x_1,\cdots,x_n)\neq (0,\cdots,0):\,\, q(x_1,\cdots,x_n)>0.$$

\item  实对称阵 $A$ 称为是正定矩阵，如果实二次型 $q=X^{\,t}AX$ 是正定二次型。
\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{9.3.2. 例子 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：}
\begin{enumerate}
\item  {\color{red}验证实二次型 $q(x_1,x_2)=x_1^2+4x_1x_2+5x_2^2$ 是不是正定的。}
\item  {\color{red}验证矩阵 $A=\begin{pmatrix} 1&2 \\ 2&5 \end{pmatrix}$ 是不是正定的。}
\end{enumerate}

\item 解答：
\begin{enumerate}
\item 是正定的。
\item 是正定的。
\end{enumerate}


\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{9.3.3. 半正定二次型和半正定矩阵的定义}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：什么是半正定二次型？}

\item  解答：

\begin{enumerate}

\item  实二次型 $q(x_1,\cdots,x_n)$ 称为是半正定的，如果对任意实数 $x_1,\cdots,x_n$, 这个二次型的值总是大于等于零的，也就是说，
$$\forall (x_1,\cdots,x_n):\,\, q(x_1,\cdots,x_n)\ge 0.$$

\item  举出一个半正定但不是正定的二次型的例子：
$$q(x_1,x_2)=x_1^2+4x_1x_2+4x_2^2. $$

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{9.3.4. 定理9.3.1. 实二次型是正定的充分必要条件1}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  {\color{red}定理：实二次型 $q(x_1,\cdots,x_n)$ 是正定的当且仅当它的秩和符号差都是 $n$. } 
\item 证明：（充分性）
\begin{enumerate}
\item  设实二次型 $q(x_1,\cdots,x_n)=X^tAX$ 的秩与符号差都是 $n$.
\item  根据秩与符号差的定义，存在实数可逆矩阵 $P$, 使得 $P^tAP=E_n$. 
\item  记 $X=PY$, 则有 $q(X)=q(PY)=(PY)^tA(PY)=y_1^2+\cdots+y_n^2$. 
\item  当 $x_1,\cdots,x_n$ 不全为零时，$y_1,\cdots,y_n$ 不全为零，故 $q(x_1,\cdots,x_n)>0$. 
\item  因此实二次型 $q(x_1,\cdots,x_n)$ 是正定的。
\end{enumerate}

\item 证明：（必要性）
\begin{enumerate}
\item  设实二次型 $q(x_1,\cdots,x_n)$ 是正定的。
\item  存在实数可逆矩阵 $P$, 使得 $P^tAP=\text{diag}\{E_p,-E_q,O\}$. 
\item  若 $p<n$, 则可以找到不全为零的实数 $x_1,\cdots,x_n$, 使得 $q(x_1,\cdots,x_n)\le 0$. 
\item  因此只能是 $p=n$, 即秩与符号差都是 $n$. 
\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{9.3.5. 顺序主子式的定义}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：什么是实对称矩阵的顺序主子式？}

\item 解答：

\begin{enumerate}

\item  一个 $n$ 阶实对称矩阵 $A$ 的 $k(1\le k\le n)$ 阶顺序主子式，是指将 $A$ 的前 $k$ 行和前 $k$ 列拿来组成的行列式。

\item  矩阵 {\footnotesize $A=\begin{pmatrix} 1&2&3 \\ 2&4&5 \\ 3&5&6 \end{pmatrix}$} 的1、2、3阶顺序主子式分别为
{\footnotesize 
$$
A_1 = 1, \,\, 
A_2 = \begin{vmatrix} 1&2 \\ 2&4 \end{vmatrix}, \,\,
A_3 = \begin{vmatrix} 1&2&3 \\ 2&4&5 \\ 3&5&6 \end{vmatrix}. 
$$
}

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{9.3.6. 定理9.3.2. 实二次型是正定的充分必要条件2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  {\color{red}定理：实二次型 $q=X^tAX$ 是正定的当且仅当对称矩阵 $A$ 的所有顺序主子式都大于零。}

\item 证明：这是本课程的几个重要定理之一。证明分两部分。

\begin{enumerate}

\item  必要性：如果实对称阵 $A$ 的某个顺序主子式小于等于零，那么实二次型 $q=X^{\,t}AX$ 不是正定的。

\item  充分性：如果实对称阵 $A$ 的所有顺序主子式都大于零，那么实二次型 $q=X^{\,t}AX$ 是正定的。

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{9.3.7. 例子 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  {\color{red}问题：用顺序主子式的方法，判断下述实二次型是不是正定的，
\begin{eqnarray*}
q(x_1,x_2,x_3) = 10x_1^2-2x_2^2+3x_3^2+4x_1x_2+4x_1x_3.
\end{eqnarray*}
}

\item 解答：相应的实对称阵为 {\footnotesize $A=\begin{pmatrix} 10&2&2 \\ 2&-2&0 \\ 2&0&3 \end{pmatrix}$}. 
计算顺序主子式，可得
{\footnotesize 
$$
A_1 = 10>0, \,\, 
A_2 = \begin{vmatrix} 10&2 \\ 2&-2 \end{vmatrix}<0. 
%A_3 = \begin{vmatrix} 10&2&2 \\ 2&-2&0 \\ 2&0&3 \end{vmatrix}. 
$$
}
因此这个实二次型不是正定的。

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{9.3.8. 必要性 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  {\color{red}问题：设实对称矩阵 $A$ 的行列式小于等于零，那么实二次型 $q(X)=X^{\,t}AX$ 不是正定的。}

\item  证明：
\begin{enumerate}
\item  因为 $A$ 是实对称矩阵，所以存在实数可逆矩阵 $P$ 使得 

\vspace{-0.4cm}
$$B=P^tAP=\text{diag}\{E_p,-E_q,O\}. $$
\vspace{-0.6cm}

\item  因为 $\det(A)\le 0$, 以及 $\det(P)\neq 0$, 所以 $p<n$.  
\item  令变量代换 $X=PY$, 则 

\vspace{-0.4cm}
$$q(PY) = Y^{\,t}BY = y_1^2+\cdots+y_p^2-y_{p+1}^2-\cdots-y_{p+q}^2. $$
\vspace{-0.6cm}

\item  因为 $p<n$, 所以存在 $Y\neq 0$ 使得 $q(PY)\le 0$. 
\item  因为 $X=PY$ 是可逆的变量代换，所以存在 $X\neq 0$ 使得 $q(X)\le 0$.  
\end{enumerate}

\vfill 

\item  注：$Y\neq 0$ 是指 $y_1,\cdots,y_n$ 不全为零。
$X\neq 0$ 是指 $x_1,\cdots,x_n$ 不全为零。

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{9.3.9. 必要性 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：设实对称阵 $A_1$ 的行列式小于等于零，设矩阵 

\vspace{-0.2cm}
{\footnotesize $$A=\begin{pmatrix} A_1&A_2 \\ A_2^t&A_3 \end{pmatrix}$$} 
\vspace{-0.2cm}

是阶数更高的实对称阵，
那么实二次型 $q(X)=X^{\,t}AX$ 不是正定的。}

\item  证明：
\begin{enumerate}
\item  设 $A_1$ 是 $m$ 阶矩阵，$A$ 是 $n$ 阶矩阵，由条件知 $m<n$. 
\item  将未知数的行向量 $X^t$ 写成 $(X_1^t,X_2^t)$ 的形式，其中 $X_1$ 的长度是 $m$. 
\item  实二次型 $q(X)=X^{\,t}AX$ 可以写成

\vspace{-0.2cm}
{\footnotesize 
$$q(X)=X^{\,t}AX = 
(X_1^t,X_2^t) \begin{pmatrix} A_1&A_2 \\ A_2^t&A_3 \end{pmatrix} \begin{pmatrix} X_1 \\ X_2 \end{pmatrix} 
= X_1^tA_1X_1 + \cdots +X_2^tA_3X_2. $$
}
\vspace{-0.2cm}

\item  因为 $\det(A_1)\le 0$, 所以存在 $X_1\neq 0$ 使得 $X_1^tA_1X_1\le 0$. 
\item  取 $X^t=(X_1^t,0)$, 则有 $X\neq 0$, 且 $X^{\,t}AX = X_1^tA_1X_1 \le 0$. 

\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{9.3.10. 充分性：二阶的情形 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  {\color{red}问题：设 \,{\footnotesize $A=\begin{pmatrix} a&b \\ b&c \end{pmatrix}$} 是一个实对称阵。如果 $A$ 的两个顺序主子式都大于零，那么实二次型 $q=X^{\,t}AX$ 是正定的。}

\item 证明：

\begin{enumerate}

\item  这时 $a$ 与 $\det(A)$ 是矩阵 $A$ 的两个顺序主子式。

\item  因为 $a>0$, 所以存在初等矩阵 $P_1$ 使得 \,{\footnotesize $P_1^tAP_1=\begin{pmatrix} 1&b_1 \\ b_1&c_1 \end{pmatrix}$}. 

\item  存在初等矩阵 $P_2$ 使得 \,{\footnotesize $P_2^tP_1^tAP_1P_2=\begin{pmatrix} 1&0 \\ 0&c_2 \end{pmatrix}$}. 

\item  因为 $\det(A)>0$, 所以 $c_2>0$. 

\item  存在初等矩阵 $P_3$ 使得 \,{\footnotesize $P_3^tP_2^tP_1^tAP_1P_2P_3 = \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix}$}. 


\end{enumerate}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{9.3.11. 充分性：三阶的情形 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  {\color{red}问题：设三阶实对称阵
\,{\footnotesize $$A=\begin{pmatrix} a&b&m \\ b&c&n \\ m&n&k \end{pmatrix} $$ } 
的三个顺序主子式都大于零，那么实二次型 $q=X^{\,t}AX$ 是正定的。}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{9.3.12. 充分性：三阶的情形的证明 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{enumerate}

\item  记 \,{\footnotesize $A_1=\begin{pmatrix} a&b \\ b&c \end{pmatrix}$}, 
则存在实数二阶可逆矩阵 $P$, 使得 
\,{\footnotesize $P^tA_1P = \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix}$}. 

\item  记 \,{\footnotesize $Q=\begin{pmatrix} P&0 \\ 0&1 \end{pmatrix}$}, 则有 
\,{\footnotesize $Q^tAQ = \begin{pmatrix} 1&0&m_1 \\ 0&1&n_1 \\ m_1&n_1&k_1 \end{pmatrix}$}. 

\item  存在初等矩阵 $Q_1,Q_2$ 使得 
\,{\footnotesize $Q_2^tQ_1^tQ^tAQQ_1Q_2 = \begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&k_2 \end{pmatrix}$}. 

\item  因为 $\det(A)>0$, 所以 $k_2>0$. 

\item  存在初等矩阵 $Q_3$ 使得 
\,{\footnotesize $Q_3^tQ_2^tQ_1^tQ^tAQQ_1Q_2Q_3 = \begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}$}. 


\end{enumerate}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{习题(9.3)\#1 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  {\color{red} 问题：判断下列实二次型是不是正定的：}
\begin{enumerate}
\item  {\color{red} $q=10x_1^2 -2x_2^2 +3x_3^2 +4x_1x_2 +4x_1x_3$. } 
\item  {\color{red} $q=5x_1^2 +x_2^2 +5x_3^2 +4x_1x_2 -8x_1x_3 -4x_2x_3$. }
\end{enumerate}

\item 思路：计算对应的实对称矩阵的顺序主子式。

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{习题(9.3)\#2 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  {\color{red} 问题：实数 $\lambda$ 取什么值时，下述实二次型是正定的？
$$q=\lambda(x_1^2 +x_2^2 +x_3^2) +2x_1x_2 -2x_1x_3 -2x_2x_3 +x_4^2. $$ 
}

\item 思路：计算对应的实对称矩阵的顺序主子式。

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{习题(9.3)\#3 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  {\color{red} 问题：设 $A$ 是实对称矩阵。证明存在实数 $T$, 使得对任意 $t\ge T$, 实对称矩阵 $A+tE$ 都是正定矩阵。
}

\item 思路：计算矩阵 $A+tE$ 的顺序主子式。

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{习题(9.3)\#4 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  {\color{red} 问题：证明正定矩阵的任意主子式的值都大于零。
}

\item 思路：例如，设实对称矩阵 \,{\footnotesize $A=\begin{pmatrix} a&b&m \\ b&c&n \\ m&n&k \end{pmatrix}$} 是正定的，则有 
{\footnotesize 
\begin{eqnarray*}
a>0, \, c>0, \, k>0, \, 
\begin{vmatrix} a&b \\ b&c \end{vmatrix}>0, \, 
\begin{vmatrix} a&m \\ m&k \end{vmatrix}>0, \, 
\begin{vmatrix} c&n \\ n&k \end{vmatrix}>0, \, 
\begin{vmatrix} a&b&m \\ b&c&n \\ m&n&k \end{vmatrix}>0. 
\end{eqnarray*}
}

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{习题(9.3)\#5 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  {\color{red} 问题：设 $A=(a_{ij})$ 是 $n$ 阶正定矩阵。证明：} 
\begin{enumerate}
\item  {\color{red} $\det(A)\le a_{11}a_{22}\cdots a_{nn}$. } 
\item  {\color{red} 若 $\det(A)= a_{11}a_{22}\cdots a_{nn}$, 则 $A$ 是对角矩阵。 } 
\end{enumerate} 

\item  思路：
先证明 $n=2$ 的情形。

当 $n=3$ 时，使用下述初等变换化为 $n=2$ 的情形，
{\footnotesize 
$$
\begin{pmatrix} E & 0 \\ -\beta^tA_1^{-1} & 1 \end{pmatrix}
\begin{pmatrix} A_1&\beta \\ \beta^t & a_{33} \end{pmatrix}
\begin{pmatrix} E& -A_1^{-1}\beta \\ 0&1 \end{pmatrix}
=\begin{pmatrix} A_1&0 \\ 0&a_{33}-\beta^tA_1^{-1}\beta \end{pmatrix}. 
$$
}

\end{itemize}

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{document}










